I am asking user to fill extra fields with custom form. And in one of the fields, I have to let user choose multiple hierarchical tags. For this, I need to pass the tags from a view to the template signup.html
from classes.Tags import Tags
from django.shortcuts import render_to_response
from allauth.socialaccount import views as signup_views
def signup_view(request):
tags = Tags()
parameters={}
all_tags = tags.get_tags()
parameters['all_tags'] = all_tags
response = signup_views.signup(request)
return response
And in urls.py, I added this line before the allauth urls include line.
url(r'^accounts/social/signup/', 'mainapp.signup_views.signup_view', name = 'account_signup'),
url(r'^accounts/', include('allauth.urls')),
What I need is that I need to add all_tags to the response so that I can access it from the template. How do I do that?
This link has some details on using your own signup form. IMO, you can define your own form (eventually with a custom widget for the tags) and use it directly, without having to mess with the view.
Otherwise, #PauloAlmeida is correct. You could inherit a new class off SignupView with something like:
class MySignupView(SignupView):
def get_context_data(self, **kwargs):
ret = super(MySignupView, self).get_context_data(**kwargs)
ret['all_tags'] = Tags.get_tags()
return ret
I'd rather use the custom form approach as it won't mess up the urls.py.
Related
Im trying to create a extra view in django admin, on the left navbar. This view will be responsible for uploading a file, which will be parsed in some function (in future i would like to render result of this parsing in admin page). This file wont be saved in database, so there wont be a model. Is there any possibility to add a view to django admin (left navbar) which dont have a model? I was reading a lot, and could find a solution. What i have done for now:
Created a class which inherits from AdminSite. I tried to implement get_app_list method, but variable self._build_app_dict(request) was empty array, and this means, method couldn't find a installed aps. I wanted to add new object to app_list variable, to render it on website.
Tried to override a admin templates, but couldnt render it. I tried to override app_index.html which i put on folder: app_name/templates/admin/app_index.html
Here is my code, which ofc doesnt work:
class MyCustomAdmin(AdminSite):
def get_app_list(self, request):
"""
Return a sorted list of all the installed apps that have been
registered in this site.
"""
app_dict = self._build_app_dict(request)
breakpoint()
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
for app in app_list:
app['models'].sort(key=lambda x: x['name'])
return app_list
def get_urls(self):
from django.conf.urls import url
urls = super(MyCustomAdmin, self).get_urls()
urls += [
url(r'^my_custom_view/$', self.admin_view(MyCustomView.as_view()))
]
return urls
class MyCustomView(View):
template_name = 'admin/app_index.html'
def get(self, request):
print('fefef')
return render(request, self.template_name, {})
def post(self, request):
pass
admin_site = MyCustomAdmin()
admin_site.get_app_list(AdminSite.get_app_list)
What I need
I'm developing a Pull Notification System to an existing Django Project. With there begin over 100+ views I'm looking to find a way to incorporate a argument(notification queryset) into all the views, as this is rendered to my base.html which I can do by passing it into a view's arguments dictionary.
Problem
I want to do this without editing all of the views as this would take a long time and would be required for all future views to include this variable.
What I've tried
Creating a template filter and pass in request.user as variable to return notification for that user. This works, however when the user selects a 'New' notification I want to send a signal back to the server to both redirect them to the notification link and change the status of viewed to 'True' (POST or AJAX). Which would required that specific view to know how to handle that particular request.
What I've considered
• Editing the very core 'View' in Django.
I've tried my best to explain the issue, but if further info is required feel free to ask.
Thanks!
models.py
class NotificationModel(models.Model):
NOTIFICATION_TYPES = {'Red flag':'Red flag'}
NOTIFICATION_TYPES = dict([(key, key) for key, value in NOTIFICATION_TYPES.items()]).items()
notification_type = models.CharField(max_length=50, choices=NOTIFICATION_TYPES, default='')
user = models.ForeignKey(User, on_delete=models.CASCADE)
created = models.DateTimeField(blank=True, null=True)
text = models.CharField(max_length=500, default='')
viewed = models.BooleanField(default=False)
views.py Example
class HomeView(TemplateView):
template_name = 'app/template.html'
def get(self, request, *args, **kwargs):
notifications = NotificationsModel.objects.filter(user=request.user)
args={'notifications':notifications}
return render(request, self.template_name, args)
def notification_handler(self, request)
if 'notification_select' in request.POST:
notification_id = request.POST['notification_id']
notification = NotificationModel.objects.get(id=notification_id)
notification.viewed = True
notification.save()
return redirect #some_url based on conditions
def post(self, request, *args, **kwargs):
notifications = self.notification_handler(request, *args, **kwargs)
if notifications:
return notifications
return self.get(self, request)
With there begin over 100+ views I'm looking to find a way to incorporate a argument(notification queryset) into all the views, as this is rendered to my base.html which I can do by passing it into a view's arguments dictionary.
You don't have to put it in the views (and actually, you shouldn't - views shouldn't have to deal with unrelated responsabilities) - you can just write a simple custom template tag to fetch and render whatever you need in your base template. That's the proper "django-esque" solution FWIW as it leaves your views totally decoupled from this feature.
Have you considered mixins ?
class NotificationMixin:
my_var = 'whatever'
class MyDjangoView(NotificationMixin,.....,View)
pass
Better, Using django builtins..
from django.views.generic import base
class NotificationMixin(base.ContextMixin):
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['notification'] = 'whatever'
return context
and use this mixin with all your views AS THE FIRST CLASS INHERITED, See this.
For your case, Typing it instead of applying it to the base class is better, The base view class is not intended to be altered, It's meant to be extended, That's why we can solve the issue by this...
from django.generic.views import View as DjangoBaseView
from .mixins import NotificationMixin
class View(NotificationMixin, DjangoBaseView):
pass
and use any IDE to change all the imported Views to your View and not the django one.
In my url.py I have urls like:
url(r'^messstellen/monatlicher_verbrauch/(?P<pk>[0-9]+)/update/$',
generic.UpdateView.as_view(
model=MonatlicherVerbrauch,
form_class=MonatlicherVerbrauchForm,
success_url=reverse('messstellen:messstellen_index'),
template_name='messstellen/monatlich_form.html',
),
now I want to let the success_url be something like:
success_url = redirect('messstellen:messtelle_detail', pk=pk)
where the pk schould be the same like in the regex pattern (?P<pk>[0-9]+)
Is there a way to do it in the url.py view?
If you don't define success_url, then Django will use your model's get_absolute_url method, which you could define as:
class MonatlicherVerbrauch(models.Model):
...
def get_absolute_url(self):
return reverse('messstellen:messtelle_detail', args=[self.pk])
If your get_absolute_url points to a different url, then I don't think it is possible to set the success_url dynamically in the urls. You will have to override the view, and define get_success_url.
class MonatlicherVerbrauchUpdateView(UpdateView):
def get_success_url(self):
return reverse('messstellen:messtelle_detail', args=[self.object.pk])
# define these attributes in the view as well, to keep urls simple
model=MonatlicherVerbrauch,
form_class=MonatlicherVerbrauchForm,
template_name='messstellen/monatlich_form.html',
Then use MonatlicherVerbrauchUpdateView in your urls instead of UpdateView.
url(r'^messstellen/monatlicher_verbrauch/(?P<pk>[0-9]+)/update/$',
MonatlicherVerbrauchUpdateView.as_view()),
The advantage of subclassing the generic view is that it separates the logic of your views from the urls.
I have two models : Advertisment and Banner
when I using "generic view" How can I Bring together at the same time
The code below bring only one Advertisment
urlpatterns = patterns('',
url(r'^(?P<pk>\d+)/$', DetailView.as_view(
model = Advertisment,
context_object_name = 'advertisment',
), name='cars-advertisment-detail'),
url(r'^$', SearchView.as_view(), name='cars-advertisment-search'),
)
Aidan's answer is good if you only want to do it for a single view, but if you want to show banners on each page automatically, you have two main options.
One is to create a template tag that renders the banner, and add this tag to your templates where you want banners to be shown.
Your tag could look like this:
#register.inclusion_tag('banner.html')
def banner_display():
random_banner = Banner.objects.order_by('?')[0]
return {'the_banner': random_banner}
Then, you would create a template that shows the banner:
<img src="{{ the_banner.url|safe }}" />
In your templates, where you need the banner, you would just say {% banner_display %}
The other option you have is to create a custom template context processor. This will inject your banner as a normal variable in all requests. This is perhaps even simpler:
def banner_display(request):
random_banner = Banner.objects.order_by('?')[0]
return {'the_banner': random_banner}
You should save this in a file and then add it to your TEMPLATE_CONTEXT_PROCESSORS setting. Now in every template you have a variable {{ the_banner }}.
You need to override the get_context_data() method of the class based view (as described in the docs).
from django.views.generic import DetailView
class YourDetailView(DetailView):
model = Advertisment
context_object_name = 'advertisment'
def get_context_data(self, *args, **kwargs):
context = super(YourDetailView, self).get_context_data(*args, **kwargs)
if 'banner_id' in self.kwargs:
context['banner'] = get_object_or_404(Banner, pk=self.kwargs['banner_id']
return context
I guess you'll need to update your url conf to include a primary key for the Banner model too.
from your_app.views import YourDetailView
url(r'^(?P<ad_pk>\d+)/(?P<banner_pk>\d+)/$', YourDetailView.as_view(), name='cars-advertisment-detail'),
I am fairly new to Django and class based forms, and I am having trouble understanding how these interact with each other. Following from the django project example, I have tried to build a "search form", which would sit on all pages of my project:
# forms.py
from django import forms
class SearchForm(forms.Form):
myquery = forms.CharField(max_length=255,label="", help_text="sq")
def __unicode__(self):
return self.myquery
# views.py
from searchapp.forms import SearchForm
from django.views.generic.edit import FormView
from django.views.generic import TemplateView
class SearchView(FormView):
template_name = 'index.html'
form_class = SearchForm
success_url = '/searchres/'
def form_valid(self, form):
thequery=form.cleaned_data.get('myquery')
return super(SearchView, self).form_valid(form)
class Meta:
abstract = True
class SearchResView(SearchView):
template_name = 'searchres.html'
#urls.py
from django.conf.urls import patterns, include, url
from django.conf import settings
from deals.views import IndexView
from searchapp.views import SearchView, SearchResView
urlpatterns = patterns('',
url(r'^index/', SearchView.as_view(),name="home"),
url(r'^searchres/', SearchResView.as_view(),name="searchresx"),
)
The plan is the start off with a simple form for user to enter the search query, and also show the input form on the results page. I have the following questions here (sorry - I am a Django newbie esp. to Class Based Views):
How does one pass data ("thequery") to the success_url? i.e I would like success_url to have access to "thequery" so that I can use something like {{thequery}} on my template tags.
Upon submitting the form(name="home"), I see POST data from the form on my firebug, but I am able to see just "myquery" rather than "thequery". How does one use get_context_data() here to add/post "thequery" variable aswell?
Finally, I was wondering if it would be possible to construct the success_url based on "thequery" string i.e something like success_url = '/searchres/?q=' + thequery
Thank you in advance - I am hoping to learn more.
I would suggest using function based views for this. If you choose to subclass a generic view you will need to dig through a lot of documentation and possibly source code, to find the right methods to override. (If you're really keen then look at the ListView class along with the get_queryset(), get() and post() methods)
A single django view will normally handle both rendering the empty form AND processing the submitted form.
So the search page (both the form and the results), live at http://your-site.com/search. Your url conf is -
urlpatterns = patterns('',
#...
(r'^search/$', 'searchapp.views.search'),
)
And your view looks something like this -
def search(request):
if request.method == 'POST':
form = SearchForm(request.POST)
if form.is_valid():
my_query = form.cleaned_data['myquery']
object_list = YourModel.objects.filter(# some operation involving my_query)
return render_to_response('search_results.html', {'object_list': object_list})
else:
form = SearchForm()
render_to_response('search_form.html', {'form': form})
(Note I've assumed your form method is post rather than get - I know this isn't great http but it's a common pattern with django)
To respond to your questions -
Don't use your own method for cleaning data. Add a clean_myquery method to your form and access it with form.fields['myquery'].clean() (or if you've called is_valid() on your form, it's accessible with just form.cleaned_data['myquery']).
You want to try and avoid passing data for processing to the template. Do as much processing as you can in the view, then render the template. However if you want to pass myquery as a string for the template to render, then add it in to the context dictionary (the second non-key-word argument) in render_to_response -
return render_to_response('search.html', {'object_list': object_list, 'myquery': my query})
The post data is constructed from the form fields. You don't have a form field thequery. The view is processing the POST data - it's not creating it that's done by the html (which in turn is constructed by the Form class). Your variable thequery is declared in the view.
Django's URL dispatcher ignores query strings in the URL so http://your_site.com/ssearch will be processed by the same view as http://your_site.com/search?myquery=findstuff. Simply change the html in the template from <form method='post'> to and access the data in django with request.GET. (You'll need to change the code from the view I described above to include a new check to see whether you're dealing with a form submission or just rendering a blank form)
Have a good read of the docs on views, forms and the url dispatcher.